An artillery shell is launched on a flat, horizontal field at an angle of α = 33.8° with respect to the horizo

Question

An artillery shell is launched on a flat, horizontal field at an angle of α = 33.8° with respect to the horizontal and with an initial speed of v0 = 256 m/s. What is the horizontal distance covered by the shell after 5.10 s of flight?
1.08×103 m (in m) answer
What is the height of the shell at this moment?
5.99×102 m (in m) answer

i know the answer but could someone show me how to work this out...need to study for final...thanks

Answer 1

d = rt
d = 256(cos33.8)(5.10)
d = 1084.93 m
s = s(zero) +v(zero)t +(1/2)gt^2 (remember g is negative)
s = 256(sin33.8)(5.10) + (1/2)(-9.8)(5.10)^2
s = 598.85 m

Answer 2

v0 in horizontal direction=256cos33.8=212.73 m/s
hor dist after 5.1 sec=v0xt=212.73x5.1=1085 m
v0 in vertical diection=256sin33.8=142.41 m/s
height at t sec=v0xt-0.5gt^2
plug in the ver v0=142.41 and t=5.1 sec
h=599m
you're right
good job